Past exam of the options I course of the University of Oxford / 2005 / C / 1 / b

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- Solution b

Solution

Euler's criterion implies multiplicativity of the symbol on nonzero residues and gives

(- 1/ p) = (- 1)^{(p - 1) /2}

. Apply quadratic reciprocity with the odd prime 3:

(\frac{- 3}{p}) = (- 1)^{(p - 1) /2} (\frac{3}{p}) = (- 1)^{(p - 1) /2} (- 1)^{(p - 1) /2} (\frac{p}{3}) = (\frac{p}{3}) .

The only nonzero square modulo three is one. Since

p > 3

,

(\frac{- 3}{p}) = {1, - 1, p \equiv 1 (mod 3), p \equiv 2 (mod 3) .

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